question let the region r be the area enclosed by the function f(x)=x³ + 1, the horizontal line y = 9, and…

Explanation:

Step1: Find the intersection point

Set $x^{3}+1 = 9$, then $x^{3}=8$, so $x = 2$.

Step2: Use the disk - washer method formula

The formula for the volume $V$ of the solid of revolution about the line $y = k$ using the disk - washer method is $V=\pi\int_{a}^{b}[(k - g(x))^{2}-(k - h(x))^{2}]dx$. Here, $k = 9$, $g(x)=x^{3}+1$, $h(x)=0$, $a = 0$, $b = 2$. So $V=\pi\int_{0}^{2}(9-(x^{3}+1))^{2}dx=\pi\int_{0}^{2}(8 - x^{3})^{2}dx$.

Step3: Expand the integrand

Expand $(8 - x^{3})^{2}=64-16x^{3}+x^{6}$.

Step4: Integrate term - by - term

$\int(64-16x^{3}+x^{6})dx=64x-4x^{4}+\frac{1}{7}x^{7}+C$.

Step5: Evaluate the definite integral

$V=\pi\left[64x-4x^{4}+\frac{1}{7}x^{7}\right]_{0}^{2}=\pi\left(64\times2-4\times2^{4}+\frac{1}{7}\times2^{7}\right)=\pi\left(128 - 64+\frac{128}{7}\right)=\pi\left(64+\frac{128}{7}\right)=\pi\times\frac{448 + 128}{7}=\frac{576\pi}{7}\approx257.831$.

Answer:

$257.831$